Answer:
The probability that he has exactly 2 hits in his next 7 at-bats is 0.3115.
Step-by-step explanation:
We are given that a baseball player has a batting average of 0.25 and we have to find the probability that he has exactly 2 hits in his next 7 at-bats.
Let X = Number of hits made by a baseball player
The above situation can be represented through binomial distribution;
[tex]P(X = r) = \binom{n}{r}\times p^{r} \times (1-p)^{n-r}; x = 0,1,2,......[/tex]
where, n = number of trials (samples) taken = 7 at-bats
r = number of success = exactly 2 hits
p = probability of success which in our question is batting average
of a baseball player, i.e; p = 0.25
SO, X ~ Binom(n = 7, p = 0.25)
Now, the probability that he has exactly 2 hits in his next 7 at-bats is given by = P(X = 2)
P(X = 2) = [tex]\binom{7}{2}\times 0.25^{2} \times (1-0.25)^{7-2}[/tex]
= [tex]21 \times 0.25^{2} \times 0.75^{5}[/tex]
= 0.3115